Question

The foot of the perpendicular from the point \( (1,6,3) \) to the line \( \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} \) is

• A. \( (1,3,5) \)
• B. \( (-1,-1,-1) \)
• C. \( (2,5,8) \)
• D. \( (-2,-3,-4) \)
• E. \( \left(\frac{1}{2},2,\frac{1}{2}\right) \)

Hint:

Foot of perpendicular uses dot product = 0 condition.

Answer 1

The Correct Option is A

Concept: Foot of perpendicular lies on line and vector from point is perpendicular to direction vector.

Step 1: Parametric form
\[ x=t,\quad y=1+2t,\quad z=2+3t \]

Step 2: Coordinates of foot
\[ F=(t,1+2t,2+3t) \]

Step 3: Direction vector
\[ \vec{d}=(1,2,3) \]

Step 4: Perpendicular condition
\[ (PF)\cdot d = 0 \] \[ (t-1,1+2t-6,2+3t-3)\cdot(1,2,3)=0 \]

Step 5: Simplify
\[ (t-1,2t-5,3t-1) \] Dot: \[ (t-1)+2(2t-5)+3(3t-1)=0 \] \[ t-1+4t-10+9t-3=0 \] \[ 14t-14=0 \Rightarrow t=1 \]

Step 6: Final point
\[ F=(1,3,5) \] \[ \boxed{(1,3,5)} \]